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MATHS M.C.Q (1 Marks)

Model Paper 1 · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 18 questions.

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M.C.Q (1 Marks)

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18 questions · self-marked practice — reveal the answer and mark yourself.

MCQ 11 Mark
The centroid of a triangle is (2, 7) and two of its vertices are (4, 8) and (-2, 6). The third vertex is
  • A
    $(0,0)$
  • B
    (4, 7)
  • C
    (7, 7)
  • D
    (7, 4)
Answer
(b) (4, 7) 
Explanation: Let A (4, 78) and B (-2, 6) be the given vertex. Let C(h, k) be the third vertex. 
The centroid of $\triangle ABC$ is $\left(\frac{4-2+h}{3}, \frac{8+6+k}{3}\right)$
It is given that the centroid of triangle ABC is (2, 7) as obtained from above formula, 
$\therefore \frac{4-2+h}{3}=2, \frac{8+6+k}{3}=7$
$\Rightarrow h=4, k=7$
Thus, the third vertex is $(4,7)$
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MCQ 21 Mark
If $\sin \theta+\operatorname{cosec} \theta=2$, then $\sin ^2 \theta+\operatorname{cosec}^2 \theta$ is equal to
  • A
    2
  • B
    1
  • C
    3
  • D
    4
Answer
(a) 2
Explanation: Given $\sin \theta+\operatorname{cosec} \theta=2$
Squaring on both sides, we get  
$\begin{array}{l}\sin ^2 \theta+\csc ^2 \theta+2 \sin \theta \csc \theta=4[\because \sin \theta \operatorname{cosec} \theta=1] \\ 
\Rightarrow \sin ^2 \theta+\operatorname{cosec}^2 \theta=4-2=2\end{array}$
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MCQ 31 Mark
Mark the correct answer for $(2-3 i)(-3+4 i)=$ ?
  • A
    (6 + 17i)
  • B
    (-6 + 17i)
  • C
    (6 - 15i)
  • D
    (6 - 17i)
Answer
(a) (6 + 17i)
Explanation: $(2-3 i)(-3+4 i)=\left(-6+8 i+9 i-12 i^2\right)=(-6+17 i+12)=(6+17 i)$
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MCQ 41 Mark
$\frac{1-\cos 2 x+\sin x}{\sin 2 x+\cos x}=?$
  • A
    cosec x
  • B
    sec x
  • C
    cot x
  • D
    tan x
Answer
(d) tan x 
Explanation: $\frac{1-\cos 2 x+\sin x}{\sin 2 x+\cos x}=\frac{2 \sin ^2 x+\sin x}{2 \sin x \cos x+\cos x}=\frac{\sin x(2 \sin x+1)}{\cos x(2 \sin x+1)}=\tan x$
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MCQ 61 Mark
  • A
    − x  > − 5
  • B
    none of these
  • C
    − x <  5
  • D
    x  > − 5
Answer
(a) − x > − 5
Explanation: Given x < 5
Multiplying both sides of the above inequality by -1,we get
− x > − 5 (The sign of the inequality is to be reversed  if both sides of an inequality are multiplied by the same negative real number)
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MCQ 71 Mark
${ }^{36} C_{34}=?$
  • A
    610
  • B
    630
  • C
    1224
  • D
    612
Answer
(b) 630 
Explanation: ${ }^{36} C _{34}={ }^{36} C _{(36-34)}={ }^{36} C _2=\frac{36 \times 35}{2}=630$
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MCQ 91 Mark
$(3+6+12+\ldots+1536)=?$
  • A
    1023
  • B
    3069
  • C
    2046
  • D
    4092
Answer
(b) 3069 
Explanation: This is a GP in which $a =3, r =\frac{6}{3}=2$ and $l =1536$
$\therefore$ required sum $=\frac{(l r-a)}{(r-1)}=\frac{(1536 \times 2-3)}{(2-1)}=(3072-3)=3069$.
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MCQ 101 Mark
If a set A has n elements then the total number of subsets of A is
  • A
    2n
  • B
    n
  • C
    $2^n$
  • D
    $n^2$
Answer
(c) $2^{ n }$
Explanation: The total no of subsets $=2^{ n }$
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MCQ 111 Mark
Domain of definition of the function $f ( x )=\frac{3}{4-x^2}+\log _{10}\left(x^3-x\right)$ is
  • A
    $(-1,0) \cup(1,2) \cup(2, \infty)$
  • B
    $(1,2) \cup(2, \infty)$
  • C
    $(-1,0) \cup(1,2)$
  • D
    $(1,2)$
Answer
(a) $(-1,0) \cup(1,2) \cup(2, \infty)$
Explanation: For f(x) to be real, we must have
$\begin{array}{l}4-x^2 \neq 0 \text { and } x^3-x>0 \\ \Rightarrow x^2 \neq 4 \text { and } x\left(x^2-1\right)>0 \\ \Rightarrow x \neq 2,-2 \text { and } x(x-1)(x+1)>0 \\ \Rightarrow x \neq 2,-2 \text { and }-1<x<0,1<x<\infty \\ \therefore \text { Domain }=(-1,0) \cup(1,2) \cup(2, \infty)\end{array}$
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MCQ 121 Mark
The radian measure of $50^{\circ} 37^{\prime} 30^{\prime \prime}$ is
  • A
    $\left(\frac{5 \pi}{16}\right)^c$
  • B
    $\left(\frac{9 \pi}{32}\right)^c$
  • C
    $\left(\frac{7 \pi}{18}\right)^c$
  • D
    $\left(\frac{11 \pi}{36}\right)^c$
Answer
(b)$\left(\frac{9 \pi}{32}\right)^c$
Explanation: $50^{\circ} 37^{\prime} 30^{\prime \prime}=50^{\circ}+\left(37 \frac{30}{60}\right)^{\prime}=50^{\circ}+\left(\frac{75}{2}\right)^{\prime}=50^{\circ}+\left(\frac{75}{2 \times 60}\right)^{\circ}=\left(50 \frac{5}{8}\right)^{\circ}=\left(\frac{405}{8}\right)^{\circ}$
$180^{\circ}=\pi^c \Rightarrow 1^{\circ}=\left(\frac{\pi}{180}\right)^c \Rightarrow\left(\frac{405}{8}\right)^{\circ}=\left(\frac{\pi}{180} \times \frac{405}{8}\right)^c=\left(\frac{9 \pi}{32}\right)^c$
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MCQ 131 Mark
The solution set for $(x+3)+4>-2 x+5$ :
  • A
    $(-\infty, 2)$
  • B
    $\left(\frac{-2}{3}, \infty\right)$
  • C
    $(-\infty,-2)$
  • D
    $(2, \infty)$
Answer
(b) $\left(\frac{-2}{3}, \infty\right)$
Explanation: $(x+3)+4>-2 x+5$
$\begin{array}{l}\Rightarrow x+7>-2 x+5 \\
\Rightarrow x+7+2 x>-2 x+5+2 x \\
\Rightarrow 3 x+7>5 \\
\Rightarrow 3 x+7-7>5-7 \\
\Rightarrow 3 x>-2 \\
\Rightarrow x>\frac{-2}{3} \\
\Rightarrow x \in\left(\frac{-2}{3}, \infty\right)\end{array}$
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MCQ 141 Mark
Let R be a relation on N defined by x + 2y = 8. The domain of R is
  • A
    (1, 2, 3, 4)
  • B
    (2, 4, 8)
  • C
    (2, 4, 6, 8)
  • D
    (2, 4, 6)
Answer
(d) (2, 4, 6)
Explanation: We have , x + 2y = 8 
$y=\frac{8-x}{2}$
since, x and y are Natural numbers, So x must be an even number. 
if x = 2, y = 3;
if x = 4, y = 2;
if x = 6, y = 1.
|So, relation R = {(2, 3), (4, 2), (6, 1)}
Hence, the domain of R is (2, 4, 6).
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MCQ 151 Mark
If $z=\left(\frac{1+i}{1-i}\right)$, then $z^4$ equals.
  • A
    $0$
  • B
    -1
  • C
    2
  • D
    1
Answer
(d) 1
Explanation:  1
$\begin{array}{l}\text { Let } z =\frac{1+i}{1-i} \\ 
z=\frac{1+i}{1-i} \times \frac{1+i}{1+i} \\ 
\Rightarrow z=\frac{1+ i ^2+2 i}{1-i^2} \\ 
\Rightarrow z=\frac{2 i}{2}\end{array}$
$\begin{array}{l}\Rightarrow z=i \\ 
\Rightarrow z^4=i^4\end{array}$
Since i2 = -1, we have: 
$\begin{array}{l}\Rightarrow z^4=i^2 \times i^2 \\ 
\Rightarrow z^4=1\end{array}$
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MCQ 171 Mark
Let A = (a, b, c), B = (a, b), C = (a, b, d), D = (c, d) and E = (d). Then which of the following statement is not correct?
  • A
    $D \supseteq E$
  • B
    $C - B = E$
  • C
    $B \cup E=C$
  • D
    $C - D = E$
Answer
(d) C - D = E 
Explanation:  C - D = (a, b, c) - (c, d) = (a, b) 
But $E=\{d\}$
Hence $C-D \neq E$
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MCQ 181 Mark
Two cards are drawn successively without replacement from a well-shuffled pack of 52 cards. The probability of drawing two aces is
  • A
    $\frac{1}{221}$
  • B
    $\frac{1}{26}$
  • C
    $\frac{1}{13}$
  • D
    $\frac{4}{223}$
Answer
(a) $\frac{1}{221}$
Explanation: Total number of ways drawing 2 cards successively without replacement 
$={ }^{52} C _1 \times{ }^{51} C _1$ and number of ways 2 aces without replacement $={ }^4 C _1 \times{ }^3 C _1$
$\therefore$ Required probability $=\frac{{ }^4 C_1 \times{ }^3 C_1}{{ }^{32} C_1 \times{ }^{51} C_1}=\frac{4 \times 3}{52 \times 51}$
$=\frac{1}{13 \times 17}=\frac{1}{22}$
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